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A topological space is noetherian if it satisfies the descending chain condition for its closed subsets. Let be $R$ a commutative ring and let $\mathrm{Spec}(R)$ its spectrum with Zariski topology.

I already know some examples of non-noetherian rings whose spectrum is noetherian, but in all these cases the spectrum is noetherian as it is finite.

Can someone give me an example of a non-noetherian ring whose spectrum is noetherian and with infinite points ?

Thanks!

Qwerty
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MG91
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2 Answers2

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Theorem: If Spec$(R)$ is Noetherian, then so is Spec$(R[X])$. [Theorem 2.5 in ``Rings with Noetherian spectrum'' by Ohm and Pendleton]

So for the example, take $R[X]$, where $R$ is any non-Noetherian ring with Noetherian spectrum.

Jason Juett
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Commutative ring without unit.

Let $\mathbb{K}$ be a field, let $R=\mathbb{K}[x_n\mid n\in\mathbb{N}]_{\displaystyle/(x_n\mid n\in\mathbb{N})^2}$: we know that $R$ is not a Noetherian ring but $\operatorname{Spec}(R)=\{(x_n\mid n\in\mathbb{N})\}$ is a Noetherian topological space.

Let $S=\bigoplus_{k\in\mathbb{N}}R$, then $\operatorname{Spec}(S)=\coprod_{k\in\mathbb{N}}\operatorname{Spec}(R)$ where the Zariski topology is the cofinite topology: $S$ is not a Noetherian ring but $\operatorname{Spec}(S)$ is a Noetherian topological space with infinitely many points.

Commutative ring with unit.

Let \begin{equation} R=\{f\in\mathbb{Q}[t]\mid f(\mathbb{Z})\subseteq\mathbb{Z}\} \end{equation} the ring of numerical polynomial (over $\mathbb{Q}$); defined: \begin{gather*} f_0=1,\,\forall n\in\mathbb{N}_{\geq1}\equiv\mathbb{N}_1,f_n=\frac{1}{n!}t(t-1)\dots(t-n+1), \end{gather*} easily one can prove that $f_n\in R$ for any $n\in\mathbb{N}_0$.

Let \begin{equation*} \forall p\in\mathbb{P}_{\geq2},\,I_p=(f_1,\dots,f_{p-1}); \end{equation*} by definition $f_p\notin I_p$, then $\{I_p\subset R\}_{p\in\mathbb{P}_{\geq2}}$ is a non stationary ascending chain of ideals in $R$, in other words $R$ is not a Noetherian ring.

Let $\mathbb{Z}[t]\stackrel{i}{\hookrightarrow}R$ the canonical inclusions, then one can consider the canonical continuous map $\operatorname{Spec}(R)\stackrel{i^{*}}{\to}\operatorname{Spec}(\mathbb{Z}[t])$. $\operatorname{Spec}(R)$ has finitely many points; because any prime ideal of $\mathbb{Z}[t]$ of the form $(t-m)$ can be extended to a prime ideal of $R$ (via $i$). By proposition V.2.7.(iii) from Cahen, Chabert - Integer-Valued Polynomials one can state that $\dim_{Krull}R=2$ and therefore $\operatorname{Spec}(R)$ is Noetherian.

Is it all clear?

Armando j18eos
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  • @MG91 I had complete my answer! – Armando j18eos Jun 29 '16 at 14:30
  • In the first paragraph, if we consider $S' = (1,1,1,…) R \oplus \bigoplus\limits_{k\in \Bbb N} R$ instead of $S$, we have a non-noetherian ring with unit. Does it however have a noetherian spectrum? – Watson Oct 08 '16 at 12:55
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    @Watson I was think to your question; $S^{\prime}$ is the (commutative) ring (with unit) of definitively constant sequnces with values in $R$, it is not Noetherian, and with an argument "à la Nagata" one can prove that $Spec S^{\prime}$ is a Noetherian topological space with infinity Krull dimension. – Armando j18eos Oct 14 '16 at 09:40